Five positive integers A, B, C, D and E
, with A < B < C < D < E
, are such that:
(i) A, B and C
(in this order) are in arithmetic sequence, and:
(ii) B, C and D
(in his order) are in geometric sequence, and:
(iii) C, D and E
(in this order) are in harmonic sequence
Determine the minimum value
such that there are precisely two quintuplets (A, B, C, D, E)
that satisfy all the given conditions.
: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.
(In reply to re: computer solutions -- no proof
by Dej Mar)
"Using substitution I was able to derive the equation,
[E-A] = (4/A)*x2 + 4*x,
such that x is the common difference between A and B. "
The above can be used to show that no further solutions exist with E-A at or below 64.
As x is the difference between B and A (the common difference of either B-A or C-B), and that difference must also be less than 64 in order for E-A to be less than or equal to 64, for a solution 4*x^2 must be less than 4*64^2 or 16,384.
But as that is the case, in order for E-A to be an integer, A must be less than 16,384. Since I checked all the values up through 200,000, all the possibilities are covered for all E-A of less than or equal to 64, and 64 is the answer.
Posted by Charlie
on 2008-10-29 11:58:22